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There is a function
\begin{align*} f(x)=x^3 + x^2 +11x +2 \\ \end{align*} Find all prime $x$ such that $f(x)$ is also a prime number.

I found that this is satisfied with an x value of 3 then the function is equal to 71, so both are primes, but I am unsure how to find other values or prove that there are no other existing solutions. I tried to use modular arithmetic, but I did not go so far.

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  • $\begingroup$ I think that, if you just worked a few values explicitly, the pattern would be obvious. $\endgroup$
    – lulu
    Jul 5, 2022 at 20:49
  • $\begingroup$ Okay. but how this could be proven without pattern recognition and stating that it is obvious. For example how to prove that there do not exist any prime values of x such that f(x) is prime also using modular arithmetic? $\endgroup$
    – user797753
    Jul 5, 2022 at 20:51
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    $\begingroup$ Of course, the proof goes by modular arithmetic. But you need to spot the pattern before it becomes clear what you want to prove. It's a really, really simple pattern. $\endgroup$
    – lulu
    Jul 5, 2022 at 20:52
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    $\begingroup$ Please change the tag to elementary-number-theory $\endgroup$
    – Bill Dubuque
    Jul 5, 2022 at 21:08
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    $\begingroup$ $\!\!\bmod 3\!:\ f(\color{#c00}{\pm1})\equiv 0\,$ so $\,f(n)\,$ is a multiple of $3$ when $n\ \rm\color{#c00}{isn't}$ (i.e. when $\,n\equiv \color{#c00}{\pm 1},\,$ e.g. all primes $\,n\neq 3)\ \ $ $\endgroup$
    – Bill Dubuque
    Jul 5, 2022 at 21:22

2 Answers 2

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Modular arithmetic is indeed the solution here. But first, we need to recall the following fact about prime numbers:

All primes $>3$ are $\pm1\pmod6$.

It's not too hard to see why. Numbers that are $0,2,4\pmod6$ are even while $3\pmod6$ is a multiple of $3$, leaving $1$ and $5$.

Now basically if $p$ is a prime that is $1\pmod6$, then $f(p)\equiv f(1)\equiv3\pmod6$, which cannot possibly be a prime (also $f(p)>3$). Similarly, if $p$ is $5\pmod6$ then $f(p)$ is also $3\pmod6$, which cannot be prime as well, so the only possible $x$ is $3$.

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Mod 3: $\mod 3$ we know by Fermat's little theorem $x^3 \equiv x\pmod 2$ and so $f(x) = x^3 + x^2 + 11x + 2 \equiv x + x^2 -x -1 \equiv x^2 - 1\equiv (x+1)(x-1)\pmod 3$.

What does this tell us?

It tells us that if $x$ is divisible by $3$ then $f(x)\equiv -1 \pmod 3$. But if $x$ is not divisible by $3$ (i.e. if $x \equiv \pm 1 \pmod 3$) then $f(x)$ is divisible by $3$ (because $x\mp 1\equiv 0 \pmod 3$).

In other words $f(x)$ is divisible by $3$ if and only if $x$ is not divisible by three.

So for $x$ and $f(x)$ to both be prime we must have exactly one of them divisible by $3$. So either $x = 3$ (the only prime divisible by $3$) and $f(x) = 71$ which just happens to be prime. Or we have $f(x)=3$ and .... $x = ????$ well, if we assume $x\in \mathbb N$ then $x \ge 1$ and $f(x) \ge 1+1+11+2 > 3$ so that is not possible.

So $x = 3$ is the only case where both are prime.

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